Bunuel
If \(|x+\frac{1}{2}|=|y+\frac{1}{2}|\), what is the value of \(x+y\)?
(1) \(xy<0\)
(2) \(x>0\) and \(y<0\)
\(|x+\frac{1}{2}|=|y+\frac{1}{2}|\)
Both sides can be \(+ve\)
Both sides can be \(-ve\)
RHS can be \(+ve \) and LHS can be \(-ve\)
RHS can be \(-ve\) and LHS can be \(+ve \)
All the above cases can be summarised in just two cases :
(i) Consider both sides \(+ve \)( this is equivalent to considering both sides negative )
(ii) Consider both sides to be of different signs.
If both sides are \(+ve\) then we get:
\(x+\frac{1}{2}= y+\frac{1}{2}\)
\(x=y \)
If both sides are of different signs then we get:
\(x+\frac{1}{2}= -y-\frac{1}{2}\)
\(x+y=-1 \)
(1) xy<0This case is not possible if \(x=y\)
Hence it must be that both the sides are of different signs and \(x+y =-1\)
SUFF.(2) x>0 and y<0Similar to statement (1) This case is not possible if \(x=y.\)
Hence it must be that both the sides are of different signs and \(x+y =-1 \)
SUFF.Ans D
Hope it helps.